$\sum_{n=1}^\infty$ $\frac{1}{\sqrt{n}} \tan(\frac{1}{n})$ How to prove  $\sum_{n=1}^\infty$ $\frac{1}{\sqrt{n}} \tan(\frac{1}{n})$ is finite. I have tried a lot. But I can not. Can anyone please help me out?
 A: For big enough $n$,
$$\frac1{\sqrt n}\tan \left(\frac1n\right)\le\frac1{\sqrt  n}\cdot\frac2n$$
Answer to OP's comment: Since 
$$\lim_{x\to0}\frac{\tan x}x=1$$
there is some $n_0\in\Bbb N$ such that for $n\ge n_0$
$$\frac{\tan\left(\cfrac1n\right)}{\cfrac 1n}<2$$
and hence,
$$\tan\left(\cfrac1n\right)<\frac2n$$
A: Use equivalents:
Near $0$, $\tan u\sim u$, so here we have $\,\tan\dfrac1n\sim_\infty \dfrac1n$ and
$$\frac1{\sqrt n}\tan\frac1n\sim_\infty \frac1{n^{3/2}}, $$
which is a convergent Riemann series.
A: 
I thought it might be of interest to proceed using trigonometric bounds introduced in elementary geometry.  To that end, we now proceed.


Note from the elementary bounds, $\sin(1/n)\le 1/n$ and $\cos(1/n)\ge \sqrt{1-(1/n)^2}$, we have for $n\ge 2$
$$0\le \frac{\tan(1/n)}{\sqrt n}\le \frac{1}{\sqrt n \sqrt{n^2-1}}$$
Furthermore, for $n\ge 2$, $n^2-1\ge \frac34 n^2$.  Therefore, we have for $n\ge 2$
$$0\le \frac{\tan(1/n)}{\sqrt n}\le \frac{2}{n^{3/2}}$$
And we are done!
A: If $f:[0,1]\to \mathbb R,$ $f(0)=0$ and $f'(0)$ exists, then
$$\sum_{n=1}^{\infty}\frac{1}{\sqrt n}f(1/n)$$
is absolutely convergent. The proof follows from applying the definition of the derivative to see $|f(x)|\le Cx$ for small positive $x.$ Hence $|f(1/n)| \le C/n$ for large $n,$ and for these $n$ the terms of the series are bounded by $Cn^{-3/2}$ in absolute value. 
A: Hint: For $0\le x\le1$,
$$
\begin{align}
\arctan(x)
&=\int_0^x\frac{\mathrm{d}t}{1+t^2}\\
&\ge\frac{x}2
\end{align}
$$
Therefore, for $0\le x\le\frac\pi4$,
$$
\tan(x)\le2x
$$

Another approach is to note that for $0\le x\le\frac\pi3$, $\cos(x)\ge\frac12$. Therefore,
$$
\begin{align}
\tan(x)
&=\frac{\sin(x)}{\cos(x)}\\
&\le2\sin(x)\\[6pt]
&\le2x
\end{align}
$$
